How to chain Python function calls so the behaviour is as follows

Question

I stumbled upon the following problem on a python challenge: Write a function that satisfies the following rule for any number of function calls.

f()()()()()(s) == 'fooooo' + s;

example:

f('it') == 'fit'; f()('x') == 'fox'; f()()('bar') == 'foobar'; f()()()('l') == 'foool'; 

The function should be stateless and should not use any variables outside the scope.

The function signature was:

def f(s=None):     # Your code here 

I thought that in order to be able to chain multiple calls we will have to return a function when no string is passed into the function, but can't figure out how to build the expected string with no external variables. Suggestions?

def f(s=None):     if s is None:         # concatenate an 'o'?         return f     else:         # Last call, return the result str.         return s 

Answer 1

An alternative to Nikola's answer is something like this:

def f(s=None):     if s: return f'f{s}'      def factory(prefix):         def inner(s=None):             return f'f{prefix}{s}' if s else factory(prefix + 'o')         return inner     return factory('o') 

using a closure and no helper function.